{"id":"cb_question_bank:b8a3b472-476e-4b02-8ee3-147abc15ccea","source_id":"cb_question_bank","source_item_id":"b8a3b472-476e-4b02-8ee3-147abc15ccea","external_id":"5c9efc7d-a1db-4027-b9b5-a10ca2abedcd","ibn":null,"question_id":"f243c383","program":"SAT","module":"math","domain_code":"S","domain":"Geometry and Trigonometry","skill_code":"S.A.","skill":"Area and volume","difficulty":"H","score_band":7,"answer_type":"mcq","answer_source":"key","layout_warning":null,"form_id":null,"module_number":null,"position":null,"stem":"Two identical rectangular prisms each have a height of 90 centimeters left parenthesis cm right parenthesis . The base of each prism is a square, and the surface area of each prism is upper K cm squared . If the prisms are glued together along a square base, the resulting prism has a surface area of StartFraction 92 Over 47 EndFraction upper K cm squared . What is the side length, in cm , of each square base?","stimulus":"","stimulus_html":null,"stem_html":"<p style=\"text-align: left;\">Two identical rectangular prisms each have a height of <math alttext=\"90 centimeters left parenthesis cm right parenthesis\"><mrow><mn>90</mn></mrow><mo>&nbsp;</mo><mtext>centimeters</mtext><mo>&nbsp;</mo><mfenced><mtext>cm</mtext></mfenced></math>. The base of each prism is a square, and the surface area of each prism is <math alttext=\"upper K cm squared\"><mi>K</mi><mo>&nbsp;</mo><msup><mtext>cm</mtext><mn>2</mn></msup></math>. If the prisms are glued together along a square base, the resulting prism has a surface area of <math alttext=\"StartFraction 92 Over 47 EndFraction upper K cm squared\"><mrow><mfrac><mn>92</mn><mn>47</mn></mfrac></mrow><mi>K</mi><mo>&nbsp;</mo><msup><mtext>cm</mtext><mn>2</mn></msup></math>. What is the side length, in <math alttext=\"cm\"><mtext>cm</mtext></math>, of each square base?</p>","rationale_html":"<p>Choice B is correct. Let <math alttext=\"x\"><mi>x</mi>\n</math> represent the side length, in <math alttext=\"cm\"><mtext>cm</mtext></math>, of each square base. If the two prisms are glued together along a square base, the resulting prism has a surface area equal to twice the surface area of one of the prisms, minus the area of the two square bases that are being glued together, which yields <math alttext=\"2 upper K minus 2 x squared cm squared\"><mn>2</mn><mi>K</mi><mo>-</mo><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>&nbsp;</mo><msup><mtext>cm</mtext><mn>2</mn></msup></math> . It’s given that this resulting surface area is equal to <math alttext=\"StartFraction 92 Over 47 EndFraction upper K cm squared\"><mfrac><mn>92</mn><mn>47</mn></mfrac><mi>K</mi><mo>&nbsp;</mo><msup><mtext>cm</mtext><mn>2</mn></msup></math>, so <math alttext=\"2 upper K minus 2 x squared equals StartFraction 92 Over 47 EndFraction upper K\"><mn>2</mn><mi>K</mi><mo>-</mo><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>=</mo><mfrac><mn>92</mn><mn>47</mn></mfrac><mi>K</mi></math>. Subtracting&nbsp;<math alttext=\"StartFraction 92 Over 47 EndFraction upper K\"><mfrac><mn>92</mn><mn>47</mn></mfrac><mi>K</mi></math> from both sides of this equation yields <math alttext=\"2 upper K minus StartFraction 92 Over 47 EndFraction upper K minus 2 x squared equals 0\"><mn>2</mn><mi>K</mi><mo>-</mo><mfrac><mn>92</mn><mn>47</mn></mfrac><mi>K</mi><mo>-</mo><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>=</mo><mn>0</mn></math>. This equation can be rewritten by multiplying <math alttext=\"2 upper K\"><mrow>\n\t<mn>2</mn>\n\t<mi>K</mi>\n</mrow>\n</math> on the left-hand side by <math alttext=\"StartFraction 47 Over 47 EndFraction\"><mfrac><mn>47</mn><mn>47</mn></mfrac></math>, which yields <math alttext=\"StartFraction 94 Over 47 EndFraction upper K minus StartFraction 92 Over 47 EndFraction upper K minus 2 x squared equals 0\"><mfrac><mn>94</mn><mn>47</mn></mfrac><mi>K</mi><mo>-</mo><mfrac><mn>92</mn><mn>47</mn></mfrac><mi>K</mi><mo>-</mo><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>=</mo><mn>0</mn></math>, or <math alttext=\"two forty sevenths upper K minus 2 x squared equals 0\"><mfrac><mn>2</mn><mn>47</mn></mfrac><mi>K</mi><mo>-</mo><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>=</mo><mn>0</mn></math>. Adding&nbsp;<math alttext=\"2 x squared\"><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup></math> to both sides of this equation yields <math alttext=\"two forty sevenths upper K equals 2 x squared\"><mfrac><mn>2</mn><mn>47</mn></mfrac><mi>K</mi><mo>=</mo><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup></math>. Multiplying both sides of this equation by <math alttext=\"StartFraction 47 Over 2 EndFraction\"><mfrac><mn>47</mn><mn>2</mn></mfrac></math> yields <math alttext=\"upper K equals 47 x squared\"><mi>K</mi><mo>=</mo><mn>47</mn><msup><mi>x</mi><mn>2</mn></msup></math>. The surface area <math alttext=\"upper K\"><mi>K</mi>\n</math>, in <math alttext=\"cm Superscript 2\"><msup><mtext>cm</mtext><mn>2</mn></msup></math>, of each rectangular prism is equivalent to the sum of the areas of the two square bases and the areas of the four lateral faces. Since the height of each rectangular prism is <math alttext=\"90 cm\"><mn>90</mn><mo>&nbsp;</mo><mtext>cm</mtext></math> and the side length of each square base is <math alttext=\"x cm\"><mi>x</mi><mo>&nbsp;</mo><mtext>cm</mtext></math>, it follows that the area of each square base is <math alttext=\"x squared cm squared\"><msup><mi>x</mi><mn>2</mn></msup><mo>&nbsp;</mo><msup><mtext>cm</mtext><mn>2</mn></msup></math> and the area of each lateral face is <math alttext=\"90 x cm squared\"><mn>90</mn><mi>x</mi><mo>&nbsp;</mo><msup><mtext>cm</mtext><mn>2</mn></msup></math>. Therefore, the surface area of each rectangular prism can be represented by the expression <math alttext=\"2 x squared plus 4 left parenthesis 90 x right parenthesis\"><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mn>4</mn><mfenced><mrow><mn>90</mn><mi>x</mi></mrow></mfenced></math>, or <math alttext=\"2 x squared plus 360 x\"><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mn>360</mn><mi>x</mi></math>. Substituting this expression for <math alttext=\"upper K\"><mi>K</mi>\n</math> in the equation <math alttext=\"upper K equals 47 x squared\"><mi>K</mi><mo>=</mo><mn>47</mn><msup><mi>x</mi><mn>2</mn></msup></math> yields <math alttext=\"2 x squared plus 360 x equals 47 x squared\"><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mn>360</mn><mi>x</mi><mo>=</mo><mn>47</mn><msup><mi>x</mi><mn>2</mn></msup></math>. Subtracting <math alttext=\"2 x squared\"><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup></math> and <math alttext=\"360 x\"><mrow>\n\t<mn>360</mn>\n\t<mi>x</mi>\n</mrow>\n</math> from both sides of this equation yields <math alttext=\"0 equals 45 x squared minus 360 x\"><mn>0</mn><mo>=</mo><mn>45</mn><msup><mi>x</mi><mn>2</mn></msup><mo>-</mo><mn>360</mn><mi>x</mi></math>. Factoring <math alttext=\"x\"><mi>x</mi>\n</math> from the right-hand side of this equation yields <math alttext=\"0 equals x left parenthesis 45 x minus 360 right parenthesis\"><mn>0</mn><mo>=</mo><mi>x</mi><mfenced><mrow><mn>45</mn><mi>x</mi><mo>-</mo><mn>360</mn></mrow></mfenced></math>. Applying the zero product property, it follows that <math alttext=\"x equals 0\"><mi>x</mi><mo>=</mo><mn>0</mn></math> and <math alttext=\"45 x minus 360 equals 0\"><mn>45</mn><mi>x</mi><mo>-</mo><mn>360</mn><mo>=</mo><mn>0</mn></math>. Adding <math alttext=\"360\"><mn>360</mn>\n</math> to both sides of the equation <math alttext=\"45 x minus 360 equals 0\"><mn>45</mn><mi>x</mi><mo>-</mo><mn>360</mn><mo>=</mo><mn>0</mn></math> yields <math alttext=\"45 x equals 360\"><mn>45</mn><mi>x</mi><mo>=</mo><mn>360</mn></math>. Dividing both sides of this equation by <math alttext=\"45\"><mn>45</mn>\n</math> yields <math alttext=\"x equals 8\"><mi>x</mi><mo>=</mo><mn>8</mn></math>. Since a side length of a rectangular prism can’t be <math alttext=\"0\"><mn>0</mn>\n</math>, the length of each square base is <math alttext=\"8\"><mn>8</mn>\n</math> <math alttext=\"cm\"><mtext>cm</mtext></math>.</p>\n<p>Choice A is incorrect and may result from conceptual or calculation errors.</p>\n<p>Choice C is incorrect and may result from conceptual or calculation errors.</p>\n<p>Choice D is incorrect and may result from conceptual or calculation errors.</p>","correct_answer":["B"],"has_media":false,"has_mathml":true,"has_table":false,"source_created_at":"2023-08-02T20:25:59.836000+00:00","source_updated_at":"2023-08-02T20:25:59.836000+00:00","options":[{"label":"A","content_html":"<p><math alttext=\"4\"><mn>4</mn>\n</math></p>","ord":0},{"label":"B","content_html":"<p><math alttext=\"8\"><mn>8</mn>\n</math></p>","ord":1},{"label":"C","content_html":"<p><math alttext=\"9\"><mn>9</mn>\n</math></p>","ord":2},{"label":"D","content_html":"<p><math alttext=\"16\"><mn>16</mn>\n</math></p>","ord":3}],"attribution":{"source_id":"cb_question_bank","source_name":"College Board digital SAT question bank (community dump)","rights_holder":"College Board","canonical_url":"https://satsuitequestionbank.collegeboard.org/","retrieved_from":"https://raw.githubusercontent.com/mdn522/sat-question-bank/main/data/cb-digital-questions.json","attribution":"SAT practice questions © College Board, from the official SAT Suite Question Bank. Retrieved via the community dump at github.com/mdn522/sat-question-bank.","disclaimer":"SAT® and PSAT/NMSQT® are trademarks registered by the College Board, which is not affiliated with, does not sponsor, and does not endorse this project.","license":"College Board copyright; source repository carries no licence. Local use only.","redistributable":false,"item_count":2017}}